[數學分析] 內積空間的不等式 Cauchy-Schwarz Inequality 與 Triangular Inequality

==============================
Theorem: (Cauchy-Schwarz Inequality) 
令 $V$ 為實數內積空間，且  ${\bf u}, {\bf v} \in V$ 則\[
|({\bf u}, {\bf v})| \le ||{\bf u}|| \; ||{\bf v}||
\]==============================

先看幾個例子
Example 1: 歐幾里德平面空間對應的 柯西不等式 
$V:=\mathbb{R}^2$ 且配備標準內積 $({\bf u}, {\bf v}) := {\bf u}^T {\bf v}$，現令 ${\bf u}:=[u_1\;\;u_2]^T; \; {\bf v}:=[v_1\;\;v_2]^T$ 則上述的 Cauchy-Schwarz Inequality 可表為
\[\begin{array}{l}
\left| {\left( {{\bf{u}},{\bf{v}}} \right)} \right| \le \left\| {\bf{u}} \right\|\left\| {\bf{v}} \right\|\\
 \Rightarrow {\left| {{{\bf{u}}^T}{\bf{v}}} \right|^2} \le \left( {{{\bf{u}}^T}{\bf{u}}} \right)\left( {{{\bf{v}}^T}{\bf{v}}} \right)\\
 \Rightarrow {\left| {{u_1}{v_1} + {u_2}{v_2}} \right|^2} \le \left( {u_1^2 + u_2^2} \right)\left( {v_1^2 + v_2^2} \right)
\end{array}\]

Example 2: 有限維歐幾里德空間對應的 柯西不等式
$V:=\mathbb{R}^n$ 且配備標準內積 $({\bf u}, {\bf v}) := {\bf u}^T {\bf v}$，現令 ${\bf u}:=[u_1,...,\;\;u_n]^T; \; {\bf v}:=[v_1,...,\;\;v_n]^T$ 則上述的 Cauchy-Schwarz Inequality 可表為
\[\begin{array}{*{20}{l}}
{\left| {\left( {{\bf{u}},{\bf{v}}} \right)} \right| \le \left\| {\bf{u}} \right\|\left\| {\bf{v}} \right\|}\\
{ \Rightarrow {{\left| {{{\bf{u}}^T}{\bf{v}}} \right|}^2} \le \left( {{{\bf{u}}^T}{\bf{u}}} \right)\left( {{{\bf{v}}^T}{\bf{v}}} \right)}\\
{ \Rightarrow {{\left| {{u_1}{v_1} + {u_2}{v_2} + ... + {u_n}{v_n}} \right|}^2} \le \left( {u_1^2 + u_2^2 + ... + u_n^2} \right)\left( {v_1^2 + v_2^2 + ... + v_n^n} \right)}
\end{array}\]

Example 3: 無窮維 實數連續函數空間 對應的 柯西不等式
$V:=C[0,1]$ 且配備內積 $(f(t), g(t)) := \int_0^1 f(t) g(t) dt$，現令 ${\bf u}:=f(t); \; {\bf v}:=g(t)$ 則上述的 Cauchy-Schwarz Inequality 可表為
\[\begin{array}{l}
\left| {\left( {f\left( t \right),g\left( t \right)} \right)} \right| \le \left\| {f\left( t \right)} \right\|\left\| {g\left( t \right)} \right\|\\
 \Rightarrow {\left| {\int_0^1 {f\left( t \right)g\left( t \right)dt} } \right|^2} \le \left( {\int_0^1 {{f^2}\left( t \right)dt} } \right)\left( {\int_0^1 {{g^2}\left( t \right)dt} } \right)
\end{array}\]

Example 4: 實數矩陣空間對應的柯西不等式
令 $V:= M_{n \times n}$ 且配備內積 $(A, B) := tr(B^T A)$ 現令 ${\bf u}:=A; \; {\bf v}:=B$ 為 $n \times n$ 矩陣，則上述的 Cauchy-Schwarz Inequality 可表為
\[\begin{array}{*{20}{l}}
{\left| {\left( {{\bf{u}},{\bf{v}}} \right)} \right| \le \left\| {\bf{u}} \right\|\left\| {\bf{v}} \right\|}\\
{ \Rightarrow {{\left| {tr\left( {{B^T}A} \right)} \right|}^2} \le tr\left( {{A^T}A} \right)tr\left( {{B^T}B} \right)}
\end{array}\]

Example 5: 隨機變數所成的 $L^2$ 空間之柯西不等式：
令 $V:= L^p :=\{X: E[|X|^2] < \infty\}$ 且配備內積 $(X,Y) := E[XY]$，其中 $X,Y$ 為隨機變數，$E[\cdot]$ 表期望值。現令 ${\bf u} := X$ 且 ${\bf v} := Y$ 則上述的 Cauchy-Schwarz Inequality 可表為
\begin{align*}
  & \left| {\left( {{\mathbf{u}},{\mathbf{v}}} \right)} \right| \leq \left\| {\mathbf{u}} \right\|\left\| {\mathbf{v}} \right\| \hfill \\
 &  \Rightarrow \left| {E\left[ {XY} \right]} \right| \leq \left\| X \right\|\left\| Y \right\| \hfill \\
  & \Rightarrow \left| {E\left[ {XY} \right]} \right| \leq \sqrt {E\left[ {{X^2}} \right]} \sqrt {E\left[ {{Y^2}} \right]}  \hfill \\
\end{align*}



Comments:
1.上述幾個例子展示了儘管所表現的樣式非常不同，但從抽象化觀點而言是同一件事情。
2. 柯西等式何時成立？

以下我們給出 Cauchy-Schwarz Inequality 的證明，此證明頗具巧思有興趣的讀者可細細品味。

Proof of Cauchy-Schwarz Inequality
令 $c \in \mathbb{R}^1$ 現在觀察
\[\begin{array}{l}
\underbrace {\left( {{\bf{u}} - c{\bf{v}},{\bf{u}} - c{\bf{v}}} \right)}_{ = {{\left\| {{\bf{u}} - c{\bf{v}}} \right\|}^2}} = \left( {{\bf{u}},{\bf{u}}} \right) + \left( { - c{\bf{v}},{\bf{u}}} \right) + \left( {{\bf{u}}, - c{\bf{v}}} \right) + \left( { - c{\bf{v}}, - c{\bf{v}}} \right)\\
 = \left( {{\bf{u}},{\bf{u}}} \right) - c\left( {{\bf{v}},{\bf{u}}} \right) - c\left( {{\bf{u}},{\bf{v}}} \right) + {c^2}\left( {{\bf{v}},{\bf{v}}} \right)\\
 = \left( {{\bf{u}},{\bf{u}}} \right) - 2c\left( {{\bf{u}},{\bf{v}}} \right) + {c^2}\left( {{\bf{v}},{\bf{v}}} \right) \;\;\;\; (*)
\end{array}\]
若 ${\bf{v}} \ne 0$ 則 $({\bf v},{\bf v})>0$ 我們可取 \[c: = \frac{{\left( {{\bf{u}},{\bf{v}}} \right)}}{{\left( {{\bf{v}},{\bf{v}}} \right)}}\]將此 $c$ 帶入 $(*)$ 可得
\[\begin{array}{l}
{\left\| {{\bf{u}} - c{\bf{v}}} \right\|^2} = \left( {{\bf{u}},{\bf{u}}} \right) - 2c\left( {{\bf{u}},{\bf{v}}} \right) + {c^2}\left( {{\bf{v}},{\bf{v}}} \right)\\
 = \left( {{\bf{u}},{\bf{u}}} \right) - 2\frac{{\left( {{\bf{u}},{\bf{v}}} \right)}}{{\left( {{\bf{v}},{\bf{v}}} \right)}}\left( {{\bf{u}},{\bf{v}}} \right) + {\left( {\frac{{\left( {{\bf{u}},{\bf{v}}} \right)}}{{\left( {{\bf{v}},{\bf{v}}} \right)}}} \right)^2}\left( {{\bf{v}},{\bf{v}}} \right)\\
 = \left( {{\bf{u}},{\bf{u}}} \right) - \frac{{{{\left( {{\bf{u}},{\bf{v}}} \right)}^2}}}{{\left( {{\bf{v}},{\bf{v}}} \right)}}
\end{array}\]但由於 ${\left\| {{\bf{u}} - c{\bf{v}}} \right\|^2} \ge 0$ 故
\[\left( {{\bf{u}},{\bf{u}}} \right) - \frac{{{{\left( {{\bf{u}},{\bf{v}}} \right)}^2}}}{{\left( {{\bf{v}},{\bf{v}}} \right)}} \ge 0 \Rightarrow \left( {{\bf{u}},{\bf{u}}} \right)\left( {{\bf{v}},{\bf{v}}} \right) \ge {\left( {{\bf{u}},{\bf{v}}} \right)^2}\]
上式結果說明在 ${\bf v} \neq 0$ 時 Cauchy-Schwarz Inequality 成立。另外我們回頭檢驗 ${\bf v} = 0$ 的情況，則此時 $\left( {{\bf{u}},{\bf{u}}} \right)\left( {{\bf{v}},{\bf{v}}} \right) \ge {\left( {{\bf{u}},{\bf{v}}} \right)^2}$ 自動滿足。故不論如何我們都有
\[{\left( {{\bf{u}},{\bf{v}}} \right)^2} \le \left( {{\bf{u}},{\bf{u}}} \right)\left( {{\bf{v}},{\bf{v}}} \right)\]至此證畢。$\square$

Comments:
1. 注意到若 ${\bf u} = c {\bf v} $ 則 Cauchy-Schwarz 等式成立。此結果背後蘊含最小平方的最佳化觀點但我們在此不作贅述。
2. 上述 Cauchy-Schwarz Inequality 可引出 Triangular Inequality

Corollary:  Triangular Inequality
令 $V$ 為實數內積空間，若 ${\bf u}, {\bf v} \in V $ 則
\[
||{\bf u} + {\bf v}|| \le ||{\bf u}|| + ||{\bf v}||
\]
Proof:
觀察
\[\begin{array}{l}
||{\bf{u}} + {\bf{v}}|{|^2} = \left( {{\bf{u}} + {\bf{v}},{\bf{u}} + {\bf{v}}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left( {{\bf{u}},{\bf{u}}} \right) + \left( {{\bf{v}},{\bf{u}}} \right) + \left( {{\bf{u}},{\bf{v}}} \right) + \left( {{\bf{v}},{\bf{v}}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = \left( {{\bf{u}},{\bf{u}}} \right) + 2\left( {{\bf{v}},{\bf{u}}} \right) + \left( {{\bf{v}},{\bf{v}}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} = {\left\| {\bf{u}} \right\|^2} + 2\left( {{\bf{v}},{\bf{u}}} \right) + {\left\| {\bf{v}} \right\|^2}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} \le {\left\| {\bf{u}} \right\|^2} + 2\left| {\left( {{\bf{v}},{\bf{u}}} \right)} \right| + {\left\| {\bf{v}} \right\|^2}
\end{array}\]由 Cauchy-Schwarz Inequality 我們有 $
|({\bf u}, {\bf v})| \le ||{\bf u}|| \; ||{\bf v}||$ 故
\[\begin{array}{l}
||{\bf{u}} + {\bf{v}}|{|^2} \le {\left\| {\bf{u}} \right\|^2} + 2\left| {\left( {{\bf{v}},{\bf{u}}} \right)} \right| + {\left\| {\bf{v}} \right\|^2}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} \le {\left\| {\bf{u}} \right\|^2} + 2||{\bf{u}}||\;||{\bf{v}}|| + {\left\| {\bf{v}} \right\|^2}\\
\begin{array}{*{20}{c}}
{}&{}&{}
\end{array} \le {\left( {\left\| {\bf{u}} \right\| + \left\| {\bf{v}} \right\|} \right)^2}
\end{array}\]對兩邊同開根號我們得到
\[
||{\bf u} + {\bf v}|| \le ||{\bf u}|| + ||{\bf v}||
\]即為所求 $\square$

Comment:
若 ${\bf u}, {\bf v}$ 互為正交，亦即 $({\bf u},{\bf v}) = 0$ 則我們有
\[
||{\bf u} + {\bf v}||^2 = ||{\bf u}||^2 + ||{\bf v}||^2
\]上述等式 可視為 畢氏定理 在向量空間中的推廣。

[線性代數] 淺論座標

令 $V$ 為 $n$ 維向量空間，則我們知道 $V$ 有基底 (basis) $S$ 且其元素為 $n$ 維向量。現在我們定義 $S :=\{{\bf v}_1,...{\bf v}_n\}$ 為向量空間 $V$ 的一組有序基底 (ordered basis) 則任意向量 ${\bf v} \in V$ 可由上述有序基底唯一表示成以下的線性組合形式：
\[
{\bf v} = a_1 {\bf v}_1 + a_2 {\bf v}_2 + ... + a_n {\bf v}_n
\]其中 $a_1,...a_n \in \mathbb{R}^1$

Definition: Coordinate Vector
定義 ${\bf v}$ 對應於有序基底 $S$ 的座標向量 (coordinate vector) 為
\[{[{\bf{v}}]_S}: = \left[ \begin{array}{l}
{a_1}\\
{a_2}\\
 \vdots \\
{a_n}
\end{array} \right]\] 且其中 $[{\bf v}]_S$ 的元素 $a_i$ 稱之為 ${\bf v}$ 對應於有序基底的座標。

Example 1:
考慮向量空間 \[
V:= P_2 := \{p(t) = a_2t^2 + a_1t + a_0: a_2,a_1,a_0 \in \mathbb{R}^2\}
\]且令基底 $S= \{t^2, t, 1\}$ 現考慮 ${\bf v}:= p(t) = \alpha t^2 + \alpha t^1 + \alpha$ 求 $[{\bf v}]_S = ?$

Solution
注意到 ${\bf v}:= p(t) = \alpha t^2 + \alpha t^1 + \alpha$，暫稱此式為 $(*)$ 又因為 ${\bf v} \in V$ 故由 $\bf v$ 可由 $S$ 的有序基底 $\{ t^2, t,1\}$ 透過線性組合唯一表示：也就是說
\[{\bf{v}} = p\left( t \right) \in {P_2} \Leftrightarrow {\bf v} = {a_2}{t^2} + {a_1}t + {a_0} \;\;\;\;\; (\star)
\]
故比較 $(*)$ 與 $(\star)$ 兩式
\[{a_2}{t^2} + {a_1}t + {a_0} = \alpha {t^2} + \alpha t + \alpha
\] 可得 $a_2 = \alpha$， $a_1 = \alpha$ 與 $a_0 = \alpha$ 故
\[{[{\bf{v}}]_S}: = \left[ \begin{array}{l}
{a_2}\\
{a_1}\\
{a_0}
\end{array} \right] = \left[ \begin{array}{l}
\alpha \\
\alpha \\
\alpha
\end{array} \right]\]

Example 2:
考慮向量空間 \[
V:= P_2 := \{p(t) = a_2t^2 + a_1t + a_0: a_2,a_1,a_0 \in \mathbb{R}^2\}
\]且令基底 $S= \{t^2-t+1, t+1, 1^2+1\}$ 現考慮 ${\bf v}:= p(t) = 4 t^2 -2 t^1 + 3$ 求 $[{\bf v}]_S = ?$

Solution:
令 \[{\left[ {\bf{v}} \right]_S} = \left[ \begin{array}{l}
{a_1}\\
{a_2}\\
{a_3}
\end{array} \right]\]且 $\bf v$ 可透過基底 $S$ 做線性組合
\[\begin{array}{l}
{\bf{v}} = {a_1}{{\bf{v}}_1} + {a_2}{{\bf{v}}_2} + {a_3}{{\bf{v}}_3}\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = {a_1}\left( {{t^2} - t + 1} \right) + {a_2}\left( {t + 1} \right) + {a_3}\left( {{t^2} + 1} \right)\\
\begin{array}{*{20}{c}}
{}&{}
\end{array} = \left( {{a_1} + {a_3}} \right){t^2} + \left( {{a_2} - {a_1}} \right)t + \left( {{a_1} + {a_2} + {a_3}} \right)\;\;\;\; (*)
\end{array}\]又因為
\[
{\bf v} = 4t^2 -2t +3 \;\;\;\; (\star)
\]故比較 $(\star)$ 與 $(*)$ 係數可得
\[\begin{gathered}
  4{t^2} - 2t + 3 = \left( {{a_1} + {a_3}} \right){t^2} + \left( {{a_2} - {a_1}} \right)t + \left( {{a_1} + {a_2} + {a_3}} \right) \hfill \\
   \Rightarrow \left\{ \begin{gathered}
  {a_1} + {a_3} = 4 \hfill \\
  {a_2} - {a_1} =  - 2 \hfill \\
  {a_1} + {a_2} + {a_3} = 3 \hfill \\
\end{gathered}  \right. \hfill \\
\end{gathered} \]
將上式改寫成矩陣求解 $a_1, a_2, a_3$如下
\[\begin{gathered}
   \left[ {\begin{array}{*{20}{c}}
  1&0&1 \\
  { - 1}&1&0 \\
  1&1&1
\end{array}} \right]\left[ \begin{gathered}
  {a_1} \hfill \\
  {a_2} \hfill \\
  {a_3} \hfill \\
\end{gathered}  \right] = \left[ \begin{gathered}
  4 \hfill \\
   - 2 \hfill \\
  3 \hfill \\
\end{gathered}  \right] \hfill \\
   \Rightarrow \left\{ \begin{gathered}
  {a_1} = 1 \hfill \\
  {a_2} =  - 1 \hfill \\
  {a_3} = 3 \hfill \\
\end{gathered}  \right. \hfill \\
\end{gathered} \]